Formatted question description: https://leetcode.ca/all/1357.html

1357. Apply Discount Every n Orders (Medium)

There is a sale in a supermarket, there will be a discount every n customer.
There are some products in the supermarket where the id of the i-th product is products[i] and the price per unit of this product is prices[i].
The system will count the number of customers and when the n-th customer arrive he/she will have a discount on the bill. (i.e if the cost is x the new cost is x - (discount * x) / 100). Then the system will start counting customers again.
The customer orders a certain amount of each product where product[i] is the id of the i-th product the customer ordered and amount[i] is the number of units the customer ordered of that product.

Implement the Cashier class:

Cashier(int n, int discount, int[] products, int[] prices) Initializes the object with n, the discount, the products and their prices.
double getBill(int[] product, int[] amount) returns the value of the bill and apply the discount if needed. Answers within 10^-5 of the actual value will be accepted as correct.

Example 1:

Input
["Cashier","getBill","getBill","getBill","getBill","getBill","getBill","getBill"]
[[3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]],[[1,2],[1,2]],[[3,7],[10,10]],[[1,2,3,4,5,6,7],[1,1,1,1,1,1,1]],[[4],[10]],[[7,3],[10,10]],[[7,5,3,1,6,4,2],[10,10,10,9,9,9,7]],[[2,3,5],[5,3,2]]]
Output
[null,500.0,4000.0,800.0,4000.0,4000.0,7350.0,2500.0]
Explanation
Cashier cashier = new Cashier(3,50,[1,2,3,4,5,6,7],[100,200,300,400,300,200,100]);
cashier.getBill([1,2],[1,2]);                        // return 500.0, bill = 1 * 100 + 2 * 200 = 500.
cashier.getBill([3,7],[10,10]);                      // return 4000.0
cashier.getBill([1,2,3,4,5,6,7],[1,1,1,1,1,1,1]);    // return 800.0, The bill was 1600.0 but as this is the third customer, he has a discount of 50% which means his bill is only 1600 - 1600 * (50 / 100) = 800.
cashier.getBill([4],[10]);                           // return 4000.0
cashier.getBill([7,3],[10,10]);                      // return 4000.0
cashier.getBill([7,5,3,1,6,4,2],[10,10,10,9,9,9,7]); // return 7350.0, Bill was 14700.0 but as the system counted three more customers, he will have a 50% discount and the bill becomes 7350.0
cashier.getBill([2,3,5],[5,3,2]);                    // return 2500.0

Constraints:

1 <= n <= 10^4
0 <= discount <= 100
1 <= products.length <= 200
1 <= products[i] <= 200
There are not repeated elements in the array products.
prices.length == products.length
1 <= prices[i] <= 1000
1 <= product.length <= products.length
product[i] exists in products.
amount.length == product.length
1 <= amount[i] <= 1000
At most 1000 calls will be made to getBill.
Answers within 10^-5 of the actual value will be accepted as correct.

Related Topics:
Design

Solution 1.

class Cashier {
    private int i;
    private int n;
    private int discount;
    private Map<Integer, Integer> d = new HashMap<>();

    public Cashier(int n, int discount, int[] products, int[] prices) {
        this.n = n;
        this.discount = discount;
        for (int j = 0; j < products.length; ++j) {
            d.put(products[j], prices[j]);
        }
    }

    public double getBill(int[] product, int[] amount) {
        int dis = (++i) % n == 0 ? discount : 0;
        double ans = 0;
        for (int j = 0; j < product.length; ++j) {
            int p = product[j], a = amount[j];
            int x = d.get(p) * a;
            ans += x - (dis * x) / 100.0;
        }
        return ans;
    }
}

/**
 * Your Cashier object will be instantiated and called as such:
 * Cashier obj = new Cashier(n, discount, products, prices);
 * double param_1 = obj.getBill(product,amount);
 */

class Cashier {
public:
    Cashier(int n, int discount, vector<int>& products, vector<int>& prices) {
        this->n = n;
        this->discount = discount;
        for (int j = 0; j < products.size(); ++j) {
            d[products[j]] = prices[j];
        }
    }

    double getBill(vector<int> product, vector<int> amount) {
        int dis = (++i) % n == 0 ? discount : 0;
        double ans = 0;
        for (int j = 0; j < product.size(); ++j) {
            int x = d[product[j]] * amount[j];
            ans += x - (dis * x) / 100.0;
        }
        return ans;
    }

private:
    int i = 0;
    int n;
    int discount;
    unordered_map<int, int> d;
};

/**
 * Your Cashier object will be instantiated and called as such:
 * Cashier* obj = new Cashier(n, discount, products, prices);
 * double param_1 = obj->getBill(product,amount);
 */

class Cashier:
    def __init__(self, n: int, discount: int, products: List[int], prices: List[int]):
        self.i = 0
        self.n = n
        self.discount = discount
        self.d = {product: price for product, price in zip(products, prices)}

    def getBill(self, product: List[int], amount: List[int]) -> float:
        self.i += 1
        discount = self.discount if self.i % self.n == 0 else 0
        ans = 0
        for p, a in zip(product, amount):
            x = self.d[p] * a
            ans += x - (discount * x) / 100
        return ans


# Your Cashier object will be instantiated and called as such:
# obj = Cashier(n, discount, products, prices)
# param_1 = obj.getBill(product,amount)

type Cashier struct {
	i        int
	n        int
	discount int
	d        map[int]int
}

func Constructor(n int, discount int, products []int, prices []int) Cashier {
	d := map[int]int{}
	for i, product := range products {
		d[product] = prices[i]
	}
	return Cashier{0, n, discount, d}
}

func (this *Cashier) GetBill(product []int, amount []int) (ans float64) {
	this.i++
	dis := 0
	if this.i%this.n == 0 {
		dis = this.discount
	}
	for j, p := range product {
		x := float64(this.d[p] * amount[j])
		ans += x - (float64(dis)*x)/100.0
	}
	return
}

/**
 * Your Cashier object will be instantiated and called as such:
 * obj := Constructor(n, discount, products, prices);
 * param_1 := obj.GetBill(product,amount);
 */

1357 - Apply Discount Every n Orders

1357. Apply Discount Every n Orders (Medium)

Solution 1.

All Problems

All Solutions

Welcome to Subscribe On Youtube

1357. Apply Discount Every n Orders (Medium)

Solution 1.

All Problems

All Solutions